Question: Brian writes down four integers $w > x > y > z$ whose sum is $44$. The pairwise positive differences of these numbers are $1, 3, 4, 5, 6,$ and $9$. What is the sum of the possible values for $w$?
The largest difference must be $w - z = 9.$  The two differences $w - x$ and $x - z$ must add up to $w - z = 9.$  Similarly, the two differences of $w - y$ and $y - z$ must add up to 9.  Thus, $\{w - x, x - z\}$ and $\{w - y, y - z\}$ must be $\{3,6\}$ and $\{4,5\}$ in some order.  This leaves $x - y = 1.$

Case 1: $\{w - x, x - z\} = \{3,6\}$ and $\{w - y, y - z\} = \{4,5\}.$

Since $w - x < w - y \le 4,$ we must have $w - x = 3,$ so $x - z = 6.$  Since $x - y = 1,$ $y - z = 5.$

Thus, $z = w - 9,$ $x = w - 3,$ and $y = w - 4.$  We also know $w + x + y + z = 44,$ so
\[w + (w - 3) + (w - 4) + (w - 9) = 44.\]Hence, $w = 15.$

Case 2: $\{w - x, x - z\} = \{4,5\}$ and $\{w - y, y - z\} = \{3,6\}.$

Since $y - z < x - z \le 4,$ we must have $y - z = 3,$ so $w - y = 6.$  Since $x - y = 1,$ $w - x = 5.$

Thus, $z = w - 9,$ $x = w - 5,$ and $y = w - 6.$   Since $w + x + y + z = 44,$
\[w + (w - 5) + (w - 6) + (w - 9) = 44.\]Hence, $w = 16.$

The sum of all possible values of $w$ is then $15 + 16 = \boxed{31}.$